Sereja and Stairs（模拟 + STL）

最新推荐文章于 2018-10-19 16:45:57 发布

原创最新推荐文章于 2018-10-19 16:45:57 发布 · 126 阅读

0 ·

CC 4.0 BY-SA版权

CF 专栏收录该内容

52 篇文章

订阅专栏

本文介绍了一个有趣的问题：如何从一组卡片中构建出最长的楼梯序列。通过使用数据结构map来跟踪每个数字的出现次数，可以有效地解决这个问题，并实现序列的最优构建。

Sereja loves integer sequences very much. He especially likes stairs.

Sequence a1, a2, ..., a|a| (|a| is the length of the sequence) is stairs if there is such index i (1 ≤ i ≤ |a|), that the following condition is met:

a1 < a2 < ... < ai - 1 < ai > ai + 1 > ... > a|a| - 1 > a|a|.

For example, sequences [1, 2, 3, 2] and [4, 2] are stairs and sequence [3, 1, 2] isn't.

Sereja has m cards with numbers. He wants to put some cards on the table in a row to get a stair sequence. What maximum number of cards can he put on the table?

Input

The first line contains integer m (1 ≤ m ≤ 105) — the number of Sereja's cards. The second line contains m integers bi (1 ≤ bi ≤ 5000) — the numbers on the Sereja's cards.

Output

In the first line print the number of cards you can put on the table. In the second line print the resulting stairs.

Sample test(s)
input
5
1 2 3 4 5
output
5
5 4 3 2 1
input
6
1 1 2 2 3 3
output
5
1 2 3 2 1

题意：

给出M（1到10^5），并给出M个数。要求对这M个数排序，排序方式为a1 < a2 < ... < ai - 1 < ai > ai + 1 > ... > a|a| - 1 > a|a|.

思路：

模拟。用 map 对这个数数值和个数进行匹配，每个数除了最大数之外，最多出现两次，分别出现在最大值两边，故 map 匹配的个数最多只为2，最大项最多为 1，最后按要求输出即可。

AC：

#include<cstdio>
#include<map>
using namespace std;
int ans[100005],k = 0;
int main()
{
    int n,sum = 0,maxnum= 0;
    map<int,int> m;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)
    {
        int num;
        scanf("%d",&num);
        if(maxnum <= num) maxnum = num;
        if(m[num] < 2)
        {
            m[num]++;
            sum++;
        }
    }    //匹配数值和个数，除最大项其他最多为2
    if(m[maxnum] != 1)
    {
        sum -= (m[maxnum] - 1);
        m[maxnum] = 1;
    }   //最大项最多只为1
    printf("%d\n",sum);
    map<int,int>::iterator it;
    for(it = m.begin();it != m.end();it++)
    {
        if(it == m.begin())   //注意输出格式
        {
            printf("%d",it -> first);
            it -> second--;
        }
        else
        {
            printf(" %d",it -> first);
            it -> second--;
        }
        if(it -> second) ans[k++] = it -> first;
    }   //前半段输出
    for(int i = k - 1;i >= 0 ;i--)
        printf(" %d",ans[i]);   //后半段输出
    printf("\n");
}